OCR FP2 2013 January — Question 1 5 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypePartial fractions with quadratic in denominator
DifficultyModerate -0.5 This is a standard partial fractions decomposition with one linear and one irreducible quadratic factor. While it's Further Maths content, the technique is routine: set up A/(x-1) + (Bx+C)/(x²+4), equate coefficients or substitute values, and solve. It requires careful algebra but no novel insight, making it slightly easier than average overall.
Spec1.02y Partial fractions: decompose rational functions

1 Express \(\frac { 5 x } { ( x - 1 ) \left( x ^ { 2 } + 4 \right) }\) in partial fractions.

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{A}{x-1} + \frac{Bx+C}{x^2+4}\)B1 Sight of expression; allow addition of constant
\(\Rightarrow 5x \equiv A(x^2+4) + (Bx+C)(x-1)\) \([+D(x-1)(x^2+4)]\)
Equate coefficients or substitute values for \(x\)M1 For equating 3 coeffs or sub 3 times
\(A = 1\)A1 For one value (not D)
\(B = -1\), \(C = 4\)A1 For 2nd and 3rd values (not D)
\(\Rightarrow \frac{5x}{(x-1)(x^2+4)} = \frac{1}{(x-1)} + \frac{4-x}{(x^2+4)}\)A1 For final answer expressed properly
[5]
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{A}{x-1} + \frac{Bx+C}{x^2+4}$ | B1 | Sight of expression; allow addition of constant |
| $\Rightarrow 5x \equiv A(x^2+4) + (Bx+C)(x-1)$ $[+D(x-1)(x^2+4)]$ | | |
| Equate coefficients or substitute values for $x$ | M1 | For equating 3 coeffs or sub 3 times |
| $A = 1$ | A1 | For one value (not D) |
| $B = -1$, $C = 4$ | A1 | For 2nd and 3rd values (not D) |
| $\Rightarrow \frac{5x}{(x-1)(x^2+4)} = \frac{1}{(x-1)} + \frac{4-x}{(x^2+4)}$ | A1 | For final answer expressed properly |
| **[5]** | | |

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1 Express $\frac { 5 x } { ( x - 1 ) \left( x ^ { 2 } + 4 \right) }$ in partial fractions.

\hfill \mbox{\textit{OCR FP2 2013 Q1 [5]}}